Problem: Factor the following expression: $x^2 - 13x + 36$
Solution: When we factor a polynomial, we are basically reversing this process of multiplying linear expressions together: $ \begin{eqnarray} (x + a)(x + b) &=& xx &+& xb + ax &+& ab \\ \\ &=& x^2 &+& {(a + b)}x &+& {ab} \end{eqnarray} $ $ \begin{eqnarray} \hphantom{(x + a)(x + b) }&\hphantom{=}&\hphantom{ xx }&\hphantom{+}&\hphantom{ (a + b)x }&\hphantom{+}& \\ &=& x^2 & & {-13}x& +& {36} \end{eqnarray} $ The coefficient on the $x$ term is $-13$ and the constant term is $36$ , so to reverse the steps above, we need to find two numbers that add up to $-13$ and multiply to $36$ You can try out different factors of $36$ to see if you can find two that satisfy both conditions. If you're stuck and can't think of any, you can also rewrite the conditions as a system of equations and try solving for $a$ and $b$ $ {a} + {b} = {-13}$ $ {a} \times {b} = {36}$ The two numbers $-4$ and $-9$ satisfy both conditions: $ {-4} + {-9} = {-13} $ $ {-4} \times {-9} = {36} $ So we can factor the expression as: $(x {-4})(x {-9})$